算法分析与设计——LeetCode:617. Merge Two Binary Trees-优快云博客

本文介绍了一种将两棵二叉树合并为一棵新二叉树的算法。当两棵树的节点重叠时，新树的节点值为两节点值之和；否则，采用非空节点作为新树的节点。通过递归方式实现该算法。

题目

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.

You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

Example 1:

Input: 
	Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
Output: 
Merged tree:
	     3
	    / \
	   4   5
	  / \   \ 
	 5   4   7

Note: The merging process must start from the root nodes of both trees.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
    }
};

思路

运用递归，如下面代码，较为简单。

代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if (!t1) return t2;
        if (!t2) return t1;

        TreeNode* node = new TreeNode(t1 -> val + t2 -> val);
        node -> left = mergeTrees(t1-> left, t2 -> left);
        node -> right = mergeTrees(t1 -> right, t2 -> right);
        return node;
    }
};